![]() ![]() Even that, although good, has it’s limits. There’s a tweak to this that improves it slightly, but a better solution is to use three added components (transistor, potentiometer and another resistor) in addition to the usual two. It’s not the best circuit, although it does provide some control. The simplest circuit uses variable resistance, adding only one component (a rheostat) to the usual two (LED and resistor). This page is somewhat specific to small white LEDs used in lighting model railroad scenery (street and building lights), but to a degree it can be applied to LEDs used for other purposes, as long as you take into account their voltage and current characteristics. But how then do you control the intensity? This page collects a variety of solutions to the problem. Another way to say this is that a LED is a current-driven device rather than a voltage-driven device. Now they are based on 10KOhms resistance.A Light Emitting Diode emits light at an intensity that depends on the current passing through the LED, not the voltage. I will improve it soon, showing the light conditions depending of the Lux value, what made it to have a more realistic results by the use of different resistances used in the circuit. I hope it could be useful to somebody else. Serial.print("Ligth conditions: ") // Print an approach to ligth conditions Serial.print(photocellReading0) // Print the measured level at pin 0 Serial.print("Voltage: ") // Print the calculated voltage returned to pin 0 Serial.print("Luminosidad 0: ") // Print the measurement (in Lux units) in the screen PhotocellReading0 = analogRead(photocellPin0) // Read the analogue pinįloat Vout0=photocellReading0*0.0048828125 // calculate the voltage We'll send debugging information via the Serial monitor Change the value of Res0 depending of what you use in the circuit you could use a double circuit (using other LDR connected to analog pin 1) to have fun testing the sensors. depending of the Resistance used, you could measure better at dark or at bright conditions. ![]() Int photocellReading0 // the analog reading from the analog resistor dividerįloat Res0=10.0 // Resistance in the circuit of sensor 0 (KOhms) Int photocellPin0 = 0 // the cell and 10K pulldown are connected to a0 Thanks to Grumpy_Mike for equations improvement. Then connect one end of a 10K resistor from Analog 0 to groundįor more information see Modified by M.A. Here is the code that i used to obtain light value in Lux units: /* Photocell simple testing sketch.Ĭonnect one end of the photocell to 5V, the other end to Analog 0. I repeated it also in Excell to be sure that i was not doing it wrong by hand or with the calculator. So i don´t know what i am doing wrong in my calculations. When i apply it to the results obtained by the circuit, i obtain a decreasing lux when i obtain a higher lecture from Arduino on pin analog 0, what it is completely wrong according to the theory of LDR sensors. Then, following the equation proposed here: Measure Light Intensity using Light Dependent Resistor (LDR), the Light measurement (in lux units) are obtained from this equation: R = 3.3 KOhms (located in the circuit in the way to the LDR) Vo = Arduino analogue lecture in Volts (ej., a lecture of 412 is equal to 2.01171875 volts, as previously explained Mike in the previous reply to this post) On the other hand, Vcc x R(light) = Vo x (R(light) + R) I assume that for most of the photocells: R(light) = 500/ Lux (KOhms) I was trying to calculate some lux values, but i have wrong results. ![]()
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